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3r^2+5r=12
We move all terms to the left:
3r^2+5r-(12)=0
a = 3; b = 5; c = -12;
Δ = b2-4ac
Δ = 52-4·3·(-12)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{169}=13$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-13}{2*3}=\frac{-18}{6} =-3 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+13}{2*3}=\frac{8}{6} =1+1/3 $
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